The length of a straight thin wire is 2 m. It is uniformly charged with a positive charge of 3 C. If charge density of the wire is 1.5 x 10-6 C/m then calculate the electric intensity due to the wire at a point 1.5 m away from the centre of the wire.
Answers
Given info : The length of a straight thin wire is 2m. it is uniformly charged and its charge density becomes 1.5 × 10¯⁶ C/m.
To find : the electric field intensity due to the wire at a point 1.5 m away from the centre of the wire is...
solution : charge density of wire, λ = 1.5 × 10¯⁶ C/m
we can apply Coulomb's law only for point charge, so here we have to first take an elementary length of wire to make it valid for coulomb's law.
so cut an element dx, x distance from the point of observation. [ see figure ]
now charge on element, dq = λdx
electric field density due to element, dE = KdQ/x²
= Kλdx/x²
= Kλ∫dx/x²
= Kλ [-1/x]
upper limit = (l + d)
lower limit = d
= Kλ [1/d - 1/(l + d)]
= Kλl/d(l + d)
now putting, K = 9 × 10^9 Nm²/C², λ = 1.5 × 10¯⁶ C/m , d = 0.5 m and l = 2m
= (9 × 10^9 × 1.5 × 10^-6 × 2)/(0.5 × 2.5)
= 12 × 10³/(1.25)
= 9600 N/C
Therefore the electric field intensity due to wire at a point 1.5m away from the centre of wire is 9600 N/C.
